2 1 Borel Regular Measures
We now state and prove an important regularity property of Borel regular outer measures:
1.1 Theorem. Suppose X is a topological space with the property that every closed Stanford Mathematics Department subset of X is the countable intersection of open sets (this trivially holds e.g. if X is Math 205A Lecture Supplement #4 a metric space), suppose is a Borel-regular outer measure on X, and suppose that X j1 1Vj , where (Vj ) < and Vj is open for each j 1; 2; : : :. Then D[ D 1 D Borel Regular & Radon Measures (1) (A) inf (U ) D U open;U A for each subset A X, and (2) (A) sup (C ) D C closed;C A for each -measurable subset A X. 1 Borel Regular Measures 1.2 Remark: In case X is a locally compact separable metric space (thus for each x X there is > 0 such that the closed ball B(x) {y X d (x; y) Recall that a Borel measure on a topological space X is a measure defined on 2 D 2 W Ä } is compact, and X has a countable dense subset), the condition X j1 1Vj the collection of Borel sets, and an outer measure on X is said to be a Borel- D[ D with Vj open and (Vj ) < is automatically satisfied provided (K) < for regular outer measure if all Borel sets are -measurable and if for each subset 1 1 A X there is a Borel set B A such that (B) (A). (Notice that this each compact K. Furthermore in this case we have from (2) above that D does not imply (B A) 0 unless A is -measurable and (A) < .) Also if n D 1 (A) sup (K) is a Borel measure on X, then we get a Borel regular outer measure on X D K compact;K A by defining (Y ) inf (B) where the inf is taken over all Borel sets B with D B Y , and this outer measure coincides with on all the Borel sets. (See for each -measurable subset A K with (A) < , because under the above 1 conditions on X any closed set C can be written C i1 1Ki , Ki compact. Q.1 of hw2.) D[ D Also, if is a Borel regular outer measure on X and if A X is -measurable Proof of 1.1. We assume first that (X) < . By Borel regularity of , for with (A) < , then we claim A is also Borel regular. Here A is the 1 1 any given A X we can select a Borel set B A with (B) (A), so it outer measure on X defined by D clearly suffices to check (1) in the special case when A is a Borel set. Now let ( A)(Y ) (A Y ): D \ To check this claim first observe that if E is -measurable and Y X is A { Borel sets A (1) holds}: D W arbitrary then ( A)(Y ) (A Y ) (A Y E) (A Y E) D \ D \ \ C \ n D Trivially A contains all open sets and one readily checks that A is closed under ( A)(Y E) ( A)(Y E), hence E is also ( A)-measurable. In \ C n both countable unions and intersections, as follows: particular all Borel sets are ( A)-measurable, so it remains to prove that for each Y X there is a Borel set B Y with ( A)(B) ( A)(Y ). To If A1; A2;::: A then for any given " > 0 there are open U1; U2;::: with 2 j D Uj Aj and (Uj Aj ) 2 ". Now one easily checks j Uj ( kAk ) prove this, first use the Borel regularity of and the fact that A is measurable n Ä [ n [ j (Uj Aj ) and j Uj ( kAk ) j (Uj Aj ) so by subadditivity we have of finite measure to pick Borel sets B1 Y A and B2 A with (B1) [ n \ n \ [ n \ D ( ( )) ( N ( )) ( (Y A) and (B A) 0, and then pick a Borel set B B A with j1 1Uj kAk < " and limN j 1Uj kAk j1 1Uj 2 3 2 [ D n [ !1 \ D n \ D \ D n \ n D n ( kAk )) < ", so both kAk and kAk are in A as claimed. (B3) 0. Then Y B1 (X A) B1 (X B2) B3 (which is a Borel D [ n [ n [ \ [ \ set) and ( A)(Y ) ( A)(B1 (X B2) B3) ((A B1) (A B3)) In particular A must also contain the closed sets, because any closed set in X Ä [ n [ D \ [ \ Ä (B1) (Y A) ( A)(Y ). can be written as a countable intersection of open sets and all the open sets D \ D Math 205A Lecture Supplement 3 4 2 Radon Measures, Representation Theorem
A A {A A X A A} A are in . Thus if we let z then z is a -algebra hence D 2 W n 2 i containing all the closed sets, and hence A contains all the Borel sets. Thus A (A ( 1 Ci`)) < 2 " z n [` contains all the Borel sets and (1) is proved in case (X) < . D 1 1 so for each i 1; 2; : : : there is a positive integer J (i) such that D To check (2) in case (X) < we can just apply (1) to X A: thus for i 1 n (A ( j J (i)Cij )) < 2 " each i 1; 2; : : : there is an open set Ui X A with (X) (A) 1=i n [j jÄ D n C D (X A) 1=i > (Ui ) (X) (Ci ), where Ci X Ui X (X A) A. Since A ( i1 1( j J (i)Cij )) i1 1(A ( j J (i)Cij ) (by De Morgan), this n C D D n n n D n \ D [j jÄ D[ D n [j jÄ Thus Ci A is closed and (Ci ) > (A) 1=i. implies (A C ) < "; In case (X) , to prove (1) we first take a Borel set B with (B) (A) n D 1 D and then apply the above result for finite measures to the Borel regular measure where C i1 1( j J (i)Cij ) is a closed subset of A. D\ D [j jÄ Vj , j 1; 2; : : :, thus obtaining, for given " > 0, open Uj B with Now define gi j J (i)Cij R by setting gi (x) (j 1)=i on Cij , j D j j W[j jÄ ! Á j j Ä ( Vj )(Uj B) < "2 and hence (Uj Vj B) "2 and by subadditivity n \ n Ä J (i). Then gi is clearly continuous and its restriction to C is continuous for this gives each i; furthermore by construction 0 f (x) gi (x) 1=i for each x C Ä Ä 2 ( (Uj Vj ) B) < ": and each i 1; 2; : : :, so that gi C converges uniformly to f C on C and hence [ \ n D j j f C is continuous. Since (Uj Vj ) is an open set containing B (hence A), this is the required j [ \ result. Similarly to check (2) in case (X) , we apply (2) for the finite measure D 1 2 Radon Measures, Representation Theorem case to the Borel regular measure Vj , giving closed Cj X such that j Cj A and ( Vj )(A Cj ) < "2 for each j 1; 2; : : :, and hence (A Vj In this section we work mainly in locally compact Hausdorff spaces, and for j n D \ n k1 1Ck ) < "2 , so by subadditivity and the fact that Vj X we have (A the reader’s convenience we recall some basic definitions and preliminary topo- [ D [ D N n ( k1 1Ck )) < " and hence (A) ( k1 1Ck ) " limN ( k 1Ck ) ". logical results for such spaces. [ D N Ä [ D C D !1 [ D C Since k 1Ck is closed for each N this is the required result. [ D Recall that a topological space is said to be Hausdorff if it has the property that for every pair of distinct points x; y X there are open sets U ; V with x U , 2 2 Using the above lemma we now prove Lusin’s Theorem: y V and U V ∅. In such a space all compact sets are automatically closed, 2 \ D the proof of which is as follows: observe that if x K then for each y K we 1.3 Theorem (Lusin’s Theorem.) Let be a Borel regular outer measure on a … 2 can (by definition of Hausdorff space) pick open Uy ; Vy with x Uy , y Vy topological space X having the property that every closed set can be expressed as the 2 2 and Uy Vy ∅. By compactness of K there is a finite set y1; : : : ; yN K with countable intersection of open sets (e.g. X is a metric space), let A be -measurable N\ D N 2 K j 1Vyj . But then j 1Uyj is an open set containing x which is disjoint with (A) < , and let f A R be -measurable. Then for each " > 0 there is [ D \ D 1 W ! from j Vyj and hence disjoint from K, so that K is closed as claimed. In fact ( ) [ a closed set C X with C A, A C < ", and f C continuous. we proved a bit more: that for each x K there are disjoint open sets U ; V n j … with x U and K V . Then if L is another compact set disjoint from K we 2 Proof: For each i 1; 2; : : : and j 0; 1; 2; : : : let can repeat this for each x L thus obtaining disjoint open U ; V with x U D D ˙ ˙ x x x 1 2 2 A f ((j 1)=i; j=i]; and K Vx, and then compactness of L implies x1; : : : ; xM L such that ij 9 2 D L M U M U M V A j 1 xj and then j 1 xj and j 1 xj are disjoint open sets containing so that Aij ; j 1; 2; : : :, are p.w.d. sets in and j1 Aij A. [ D [ D \ D D [ D 1 D L and K respectively. By a simple inductive argument (left as an exercise) we A By the remarks preceding Theorem 1.1 we know that is a Borel regular can extend this to finite pairwise disjoint unions of compact subsets: outer measure and since it is finite we can apply Theorem 1.1 to it, and hence for given " > 0 there are closed sets Cij in X with Cij Aij with ( A)(Aij 2.1 Lemma. Let X be a Hausdorff space and K1;:::; KN be pairwise disjoint i j 1 i j 1 n Cij ) (Aij Cij ) < 2 j j ", hence (Aij ( `1 Ci`)) < 2 j j " and compact subsets of X. Then there are pairwise disjoint open subsets U1;:::; UN D n n [ D 1 Math 205A Lecture Supplement 5 6 2 Radon Measures, Representation Theorem with Kj Uj for each j 1; : : : ; N . zero (the value of f X V ) on the overlap set V V V (X V ). Finally D j 1n n Á \ n we let f 2 min{f1; 2 } and observe that f is then identically 1 in the set Notice in particular that we have the following corollary of Lemma 2.1: Á 1 where f1 > 2 , which is an open set containing K, and f evidently has all the 2.2 Corollary. A compact Hausdorff space is normal: i.e. given closed disjoint remaining stated properties. subsets K1; K2 of a compact Hausdorff space, we can find disjoint open U1; U2 The following corollary of Lemma 2.4 is important: with Kj Uj for j 1; 2. D 2.5 Corollary (Partition of Unity.) If X is a locally compact Hausdorff space, Most of the rest of the discussion here takes place in locally compact Hausdorff K X is compact, and if U1;:::; UN is any open cover for K, then there exist space: A space X is said to be locally compact if for each x X there is a 2 continuous 'j X [0; 1] such that spt 'j is a compact subset of Uj for each j , neighborhood Ux of x such that the closure U x of Ux is compact. PN W ! and 'j 1 in a neighborhood of K. j 1 Á An important preliminary lemma in such spaces is: D Proof: By Lemma 2.3, for each x K there is a j {1; : : : ; N } and a neigh- 2.3 Lemma. If X is a locally compact Hausdorff space and V is a nhd. of a point 2 2 borhood Ux of x such that U x is a compact subset of this Uj . By compactness x, then there is a nhd. Ux of x such that U x is a compact subset of V . of K we have finitely many of these neighborhoods, say Ux1 ;:::; UxN with N K i 1Uxi . Then for each j 1; : : : ; N we define Vj to be the union of Proof: First pick a neighborhood U0 of x such that U 0 is compact and define [ D D all Uxi such that U xi Uj . Then the V j is a compact subset of Uj for each W U0 V . Then W is compact and hence so is the closed subset W W . D \ n j , and the Vj cover K. So by Lemma 2.4 for each j 1; : : : ; N we can select Then W W and {x} are disjoint compact sets so by Lemma 2.1 we can find D n j X [0; 1] with j 1 on V j and j 0 on X Wj for some open Wj disjoint open U1; U2 with x U1 and W W U2. Without loss of generality W ! Á Á n 2 n with W j a compact subset of Uj and Wj V j . We can also use Lemma 2.4 to we can assume U1 W (otherwise replace U1 by U1 W ). Then U 1 \ select '0 X [0; 1] with '0 0 in a neighborhood of K and '0 1 outside X U2 X (W W ) and hence U 1 W . Thus the lemma is proved with W ! N Á PN Á n n n a compact subset of j 1Vj . Then by construction i 0 i > 0 everywhere Ux U1. [ D D D on X, so we can define continuous functions 'j by j Remark: In locally compact Hausdorff space, using Lemmas 2.1 and 2.3 it is 'j ; j 1; : : : ; N : D PN D easy to check that we can select the Uj in Lemma 2.1 above to have compact i 0 i D pairwise disjoint closures. Evidently these functions have the required properties.
The following lemma is a version of the Urysohn lemma valid in locally com- We now give the definition of Radon measure. Radon measures are typically pact Hausdorff space: used only in locally compact Hausdorff space, but the definition and the first two lemmas following it are valid in arbitrary Hausdorff space: 2.4 Lemma. Let X be a locally compact Hausdorff space, K X compact, and K W , W open. Then there is an open V K with V W , V compact, and an 2.6 Definition: Given a Hausdorff space X, a “Radon measure” on X is an f X [0; 1] with f 1 in a neighborhood of K and spt f V . outer measure on X having the 3 properties: W ! Á is Borel regular and (K) < compact K X (R1) Proof: By Lemma 2.3 each x K has a neighborhood Ux with U x W 1 8 2 N and U x compact. Then by compactness of K we have K V j 1Uxj (A) inf (U ) for each subset A X (R2) N Á[ D D U open;U A for some finite collection x1; : : : ; xN K and V U x W . Now V 2 D[j 1 j is compact, so by Corollary 2.2 it is a normal space andD the Urysohn lemma (U ) sup (K) for each open U X: (R3) D K compact;K U can be applied to give f0 V [0; 1] with f0 1 on K and and f0 0 on W ! Á Á V V . Then of course the function f1 defined by f1 f0 on V and f1 0 on Such measures automatically have a property like (R3) with an arbitrary - n Á Á X V is continuous (check!) because f V is continuous and f is identically measurable subset of finite measure: n j Math 205A Lecture Supplement 7 8 2 Radon Measures, Representation Theorem
2.7 Lemma. Let X be a Hausdorff space and a Radon measure on X. Then (V K1) (K2) ". Then n Ä C automatically has the property (V U ) (V U ) (V K1) (K1) " n C \ Ä n C C (K2) (K1) 2" (A) sup (K) Ä C C D K A; K compact (K2 K1) 2" (by (i)) D [ C ((V K1) K1) 2" for every -measurable set A X with (A) < . Ä n [ C 1 (V ) 2" (Y ) 3"; D C Ä C hence (Y U ) (Y U ) (V U ) (V U ) (Y ) 3" which by Proof: Let " > 0. By definition of Radon measure we can choose an open U n C \ Ä n C \ Ä C arbitrariness of " gives (Y U ) (Y U ) (Y ), which establishes the containing A with (U A) < ", and then a compact K U with (U K) < " n C \ Ä n n -measurability of U . Thus all open sets are -measurable, and hence all Borel and finally an open W containing U A with (W (U A)) < " (so that n n n sets are -measurable, and so (R1) is established. (W ) " (U A) < 2"). Then we have that K W is a compact subset of Ä C n n U W , which is a subset of A, and also n The following lemma guarantees the convenient fact that, in a locally compact (A (K W )) (U (K W )) (U K) (W ) 3"; n n Ä n n Ä n C Ä space such that all open subsets are -compact, all locally finite Borel regular which completes the proof. outer measures are in fact Radon measures.
2.9 Lemma. X The following lemma asserts that the defining property (R1) of Radon mea- Let be a locally compact Hausdorff space and suppose that each sures follows automatically from the remaining two properties ((R2) and (R3)) open set is the countable union of compact subsets. Then any Borel regular outer X in case is finite and additive on finite disjoint unions of compact sets. measure on which is finite on each compact set is automatically a Radon mea- sure. 2.8 Lemma. Let X be a Hausdorff space and assume that is an outer measure Proof: First observe that in a Hausdorff space X the statement “each open on X satisfying the properties (R2), (R3) above, and in addition assume that set is the countable union of compact subsets” is equivalent to the statement “X is -compact (i.e. the countable union of compact sets) and every closed (K1 K2) (K1) (K2) < ; K1; K2 compact, K1 K2 ∅. [ D C 1 \ D set is the countable intersection of open sets” as one readily checks by using De Morgan’s laws and the fact that a set is open if and only if its complement Then is Borel regular, hence (R1) holds, hence is a Radon measure. is closed. Thus we have at our disposal the facts that X is -compact and every closed set is a countable intersection of open sets. The latter fact enables us to apply Theorem 1.1 (1), and we can therefore assert that Proof: Note that (R2) implies that for every set A X we can find open sets Uj such that A j Uj and (A) ( j Uj ). So to complete the proof (A) inf (U ) whenever A X has the property(1) D U open;A U \ D \ of (R1) we just have to check that all Borel sets are -measurable; since the - that open Vj with A j Vj and (Vj ) < j . 9 [ 1 8 measurable sets form a -algebra and the Borel sets form the smallest -algebra Also, by applying Theorem 1.1 (2) to the finite Borel regular measure A, which contains all the open sets, we thus need only to check that all open sets (2) (A) sup (C ); provided A is -measurable and (A) < . are -measurable. D C closed;C A 1 Let " > 0 be arbitrary, Y an arbitrary subset of X with (Y ) < and let U Now observe that by the first part of the conclusion in Lemma 2.4 there is an 1 be an arbitrary open subset of X. By (R2) we can pick an open set V Y open set V K such that V (the closure of V ) is compact. So since we are with (V ) < (Y ) " and by (R3) we can pick a compact set K1 V U given X j1 1Kj , where each Kj is compact, we can apply this with Kj in C \ D[ D with (V U ) (K1) ", and then a compact set K2 V K1 with place of K, and we deduce that there are open sets Vj in X such that j Vj X \ Ä C n [ D Math 205A Lecture Supplement 9 10 2 Radon Measures, Representation Theorem and (Vj ) < for each j , and so in this case (when X is -compact) the one can check using the dominated convergence theorem and the fact that 1 identity in (1) holds for every subset A X; that is both f and f can be expressed as the pointwise limits of increasing se- C quences of non-negative simple functions), so we can pick a simple function (A) inf (U ) for every A X; PN D U open;A U ' j 1aj Aj , where the aj are distinct non-zero reals and Aj are pair- D D wise disjoint -measurable subsets of X, such that f ' p < ". Since which is the property (R2). Next we note that if A X is -measurable, k k ' p ' f p f p < we must then have (Aj ) < for each j . then we can write A j Aj , where Aj A Kj (because X j Kj ) and k k Ä k k C k k 1 1 D[ D \ D[ Pick M > max{ a1 ;:::; aN } and use Lemma 2.7 to select compact Kj Aj (Aj ) (Kj ) < for each j , so (2) actually holds for every -measurable j j p j pj 1 p Ä 1 with (Aj Kj ) < " =(2 C M N ). Also, using the definition of Radon A in case X is -compact (i.e. in case X j1 1Kj with Kj compact), and for n p p 1 p D[ D measure, we can find open Uj Kj with (Uj Kj ) < " =(2 C M N ) any closed set C we can write C j Cj where Cj is the increasing sequence n D[ j and by Lemma 2.7 we can assume without loss of generality that these open of compact sets given by Cj C ( i 1Ki ) and so (C ) limj (Cj ) and 0 D \ [ D D sets U1;:::; UN are pairwise disjoint (otherwise replace Uj by Uj Uj , where hence (C ) supK C;K compact (K). Thus in the -compact case (2) actually 0 0 0 \ D U1 ;:::; UN are pairwise disjoint open sets with Kj Uj ). By Lemma 2.4 we tells us that (A) supK A; K compact (K) for any -measurable set A. This D have gj Cc (X) with gj aj on Kj , {x gj (x) 0} contained in a compact in particular holds for A an open set, which is the remaining property (R3) 2 Á W ¤ D subset of Uj , and sup gj aj , and hence by the pairwise disjointness of the we needed. Pj Nj Ä j j Uj we have that g j 1gj agrees with ' on each Kj and sup g sup ' < Á D j j D j j The following result is one of the main theorems related to Radon measures, M . Then ' g vanishes off the set j ((Uj Kj ) (Aj Kj )) and we have R p P R [ pn [ npP ' g d ' g d (2M ) ((Aj Kj ) asserting that for a Radon measure on a locally compact Hausdorff space, the X j (Uj Kj ) (Aj Kj ) j p j j Äp n [ j j Ä n C continuous functions with compact support are dense in L (), 1 p < . (Uj Kj )) " , and hence f g p f ' p ' g p 2", as required. Ä 1 n Ä k k Ä k k Ck k Ä Here and subsequently we use the notation We now state the Riesz representation theorem for non-negative functionals Cc (X) {f X R f is continuous with sptf compact}; on the space K , where, here and subsequently, K denotes the set of non- D W ! W C C negative Cc (X; R) functions, i.e. the set of continuous functions f X where sptf support of f closure of {x X f (x) 0}. W ! D D 2 W ¤ [0; ) with compact support. 1 2.10 Theorem. Let X be a locally compact Hausdorff space, a Radon measure p 2.12 Theorem (Riesz for non-negative functionals.) Suppose X is a locally on X and 1 p < . Then Cc (X) is dense in L (); that is, for each " > 0 and pÄ 1 compact Hausdorff space, K [0; ) with (cf ) c(f ), (f g) each f L there is a g Cc (X) such that g f p < ". W C ! 1 D C D 2 2 k k (f ) (g) whenever c 0 and f; g K , where K is the set of all non- C 2 C C In view of Lemma 2.9 and the fact that to every Borel measure on a topolog- negative continuous functions f on X with compact support. Then there is a R ical space X (i.e. every map {all Borel sets of X} [0; ] with (∅) 0 Radon measure on X such that (f ) X f d for all f K . P W ! 1 D D 2 C and ( j1 1Bj ) j1 1 (Bj ) for every pairwise disjoint collection of Borel [ D D D Before we begin the proof of 2.12 we the following preliminary observation: sets B1; B2;:::), there is a Borel regular outer measure on X defined by (A) infB Borel;B A (B), we see that Theorem 2.10 directly implies the 2.13 Remark: Observe that if f; g K with f g then g f K and D 2 C Ä 2 C following important corollary: hence (g) (f (g f )) (f ) (g f ) (f ), so D C D C 2.11 Corollary. X ( ) f; g K with g 1 on spt f If is a locally compact Hausdorff space such that every open set 2 C Á ) (f ) (sup f ) (g); f K ; spt f K: in X is the countable union of compact sets, and if is any Borel measure on X Ä 2 C 1 which is finite on each compact set, then the space Cc (X) is dense in L () and because f g f and f (sup f )g. Á Ä is the restriction to the Borel sets of a Radon measure . Proof of Theorem 2.12: For U X open, we define Proof of Theorem 2.10: Let f X be -measurable with f < R p (1) (U ) (f ); W ! p k k 1 sup and let " > 0. Observe that the simple functions are dense in L () (which D f K ;f 1;spt f U 2 C Ä Math 205A Lecture Supplement 11 12 2 Radon Measures, Representation Theorem and for arbitrary A X we define On the other hand g1 g2 1 on X (because V1 V2 ∅) and so C Ä \ D (2) (A) inf (U ): (g g ) (f ) ( ): D U A;U open sup 1 2 sup U gj K ;spt gj Vj ;gj 1;j 1;2 C Ä f K ;spt f U ;f 1 D Notice that these definitions are consistent when A is itself open, and of course 2 C Ä D 2 C Ä Hence we have proved that (K1) (K2) (U ), and taking inf over all the definitions (1),(2) guarantee (∅) 0 and that is monotone—i.e. C Ä D open U K1 K2 we have by (2) that (K1) (K2) (K1 K2), and of (3) A B (A) (B): [ C Ä [ ) Ä course the reverse inequality holds by subadditivity of , hence the hypotheses Also if f K with f 1 and V is open with V spt f then by (1) of Lemma 2.8 are all established and is a Radon measure. 2 C Ä (f ) (V ), and hence, taking inf over such V and using (2), we see Ä Next observe that by (4) we have (h) (spt h) sup h; h K , and hence (4) f K with f 1 (f ) (spt f ); Ä 2 C 2 C Ä ) Ä (h) lim (max{h 1=n; 0}) ({x h(x) > 0}) sup h; h K D n Ä W 2 and then for any open U we can use (1) and (4) to conclude !1 since h is the uniform limit of max{h 1=n; 0} in X and spt max{h 1=n; 0} (5) (U ) sup (spt f ); D f K ;f 1;spt f U {x h(x) > 0} for each n. For f K (f not identically zero) and " > 0, 2 C Ä W 2 C Notice next that if K is compact then, by Lemma 2.4, if W K is open there is we let M sup f can select points 0 t0 < t1 < t2 < : : : < tN 1 < M < tN D D 1 g K with g 1 in a neighborhood V of K and with g 1 and spt g W . with tj tj 1 < " for each j 1; : : : ; N and with ({f {tj }}) 0 for 2 C Á Ä D D Then by (3),(1) and ( ) we have, for any such g, each j 1; : : : ; N . Notice that the latter requirement is no problem because D ({f 1{t}}) 0 for all but a countable set of t > 0, by virtue of the fact that (6) (K) (V ) sup (f ) (g) (W ): D Ä D f K ;f 1;spt f V Ä Ä {x X f (x) > 0} (spt f ) < . 2 C Ä 2 W Ä 1 1 To prove that is an outer measure it still remains to check countable subad- Now let Uj f {(tj 1; tj )}, j 1; : : : ; N . (Notice that then the Uj are D D ditivity. To see this, first let U1; U2;::: be open and U j Uj , then for any pairwise disjoint and each Uj K, where K, compact, is the support of f .) D[ f K with sup f 1 and spt f U we have, by compactness of spt f , Now by the definition (1) we can find gj K such that gj 1, spt gj Uj , 2 C N Ä 2 C Ä that spt f j 1Uj for some integer N , and by using a partition of unity and (gj ) (Uj ) "=N . Also for any compact Kj Uj we can construct a [ D '1;:::;'N for spt f subordinate to U1;:::; UN (see Corollary 2.5), we have function hj K with hj 1 in a neighborhood of Kj spt gj , spt hj Uj , PN PN 2 C Á [ (f ) j 1('j f ) j 1(Uj ). Taking sup over all such f we then have and hj 1 everywhere. Then hj gj , hj 1 everywhere and spt hj is a D P D Ä D Ä Ä (U ) j1 1(Uj ). It then easily follows by applying definitions (1),(2) that compact subset of Uj and so Ä DP ( j Aj ) j (Aj ). So indeed is an outer measure on X. [ Ä (9) (Uj ) "=N (gj ) (hj ) (Uj ); j 1; : : : ; N : Finally we want to show that is a Radon measure. For this we are going Ä Ä Ä D to use Lemma 2.8 above, so we have to check the hypotheses of Lemma 2.8. Since is a Radon measure, we can in fact choose the compact Kj Uj such PN Hypothesis (R2) needed for Lemma 2.8 is true by definition and (R3) is true that (Uj Kj ) < "=N . Then, because {x (f f j 1hj )(x) > 0} n W D (Uj Kj ), by (8) we have by (5). Since we also have finiteness of (K) for compact K by (6), it remains [ n only to prove the additivity property PN PN (10) (f f j 1hj ) M j 1(Uj Kj ) "M : D Ä D n Ä (7) K1; K2 disjoint compact sets in X (K1 K2) (K1) (K2): ) [ D C Then by using (9); (10) and the linearity of (together with the fact tj 1hj U K K Ä To check this, let be any open set containing 1 2 and use Corollary 2.2 f hj tj hj ) for each j 1; : : : ; N ), we see that [ Ä D to choose disjoint open Vj Kj with Vj U , j 1; 2. Then by (3) and (1) PN P P D j 1tj 1(Uj ) "M (f j hj ) (f ) (f j hj ) "M D Ä Ä Ä C (K1) (K2) (V1) (V2) sup ((g1) (g2)) PN C Ä C D C j 1tj (Uj ) "M : gj K ;spt gj Vj ;gj 1;j 1;2 Ä C 2 C Ä D D Since trivially sup (g1 g2) D gj K ;spt gj Vj ;gj 1;j 1;2 C PN R PN 2 C Ä D j 1tj 1(Uj ) X f d j 1tj (Uj ); D Ä Ä D Math 205A Lecture Supplement 13 14 2 Radon Measures, Representation Theorem we then have sup over all such f we have (g h) (g) (h). Therefore we have C Ä C PN (g h) (g) (h) as claimed. Thus satisfies the conditions of the " (K) M j 1(tj tj 1)(Uj ) "M C D C C Ä D Theorem 2.12, hence there is a Radon measure on X such that R f d (f ) Ä X Z PN (t t )(U ) "M (h) h d; h K ; j 1; : : : ; n: j 1 j j 1 j D X 2 C D Ä D C "((K) M ); That is, we have Ä C Z where K spt f . This completes the proof of 2.12. () sup L(f ) h d; h K : D n D 2 C f Cc (X;R ); f h X n 2 j jÄ We can now state the Riesz Representation Theorem. In the statement, Cc (X; ) R Thus if j {1; : : : ; n} we have in particular (since fej f K for any n will denote the set of vector functions f X R which are continuous and 2 j j D j j 2 C W ! f Cc (X; R)) that which have compact support. (That is f 0 outside a compact subset of X.) 2 Z Á L(fej ) f d f L1() f Cc (X; R): 2.14 Theorem (Riesz Representation Theorem.) Suppose X is a locally com- j j Ä X j j Á k k 8 2 1 n Thus Lj (f ) L(fej ) extends to a bounded linear functional on L (). In pact Hausdorff space, and L Cc (X; R ) R is linear with Á W ! either of the 3 cases (i) K compact is given and we use Riesz Representation sup L(f ) < whenever K X is compact. 1 n 1 Theorem for L ( K), or (ii) L ( (X)) < and we use Riesz Repre- f Cc (X;R ); f 1;spt f K 2 j jÄ k k D 1 sentation Theorem for the finite measure case, or (iii) X j1 1Kj with Kj Then there is a Radon measure on X such that for each compact K X there D[ D n compact for each j and we use Riesz Representation Theorem for the -finite is a vector function X R with 1 everywhere and j -measurable, W ! j j D case, we know that there is a bounded -measurable function j such that j 1; : : : ; n, and with Z D Z n L(fej ) f j d; f Cc (X; R); L(f ) f d for any f Cc (X; R ) with spt f K: D X 2 D 2 X where in case (i) we impose the additional restriction spt f K. Since any In the cases when X is -compact (i.e. compact K1; K2;::: with X j Kj ) or Pn 9 D[ f (f1; : : : ; fn) can be expressed as f j 1fj ej , we thus deduce L is bounded (i.e. sup n L(f ) < ), can be chosen independent D D D f Cc (X;R ); f 1 Z 2 j jÄ j j 1 n of K. ( ) L(f ) f d; f Cc (X; R ); D X 2 ( ) ( ) Proof: We first define where 1; : : : ; n , and so by Z D Z Z (h) sup L(f ) h d sup f d sup gf d D n D n D n f Cc (X;R ); f h X f Cc (X;R ); f h X f Cc (X;R ); f 1;g K ;g h X 2 j jÄ 2 j jÄ 2 j jD 2 C Ä for any h K . We claim that has the linearity properties of Lemma 2.12. for every h K , where in case (i) we assume spt h K. Now f f 2 C 2 C j j Ä j jj j Indeed it is clear that (ch) c(h) for any constant c 0 and any h K . so we have D n 2 C Z Z Z Now let g; h K , and notice that if f1; f2 Cc (X; R ) with f1 g and 2 C 2 j j Ä sup gf d sup g d h d n f2 h, then f1 f2 g h and hence (g h) L(f1) L(f2). Taking f Cc (X;R ); f 1;g K ;g h X Ä g K ;g h X j j D X j j j j Ä j C j Ä C C C 2 j jD 2 C Ä 2 C Ä sup over all such f1; f2 we then have (g h) (g) (h). To prove the 1 Since Cc (X) is dense in L (), we can choose a sequence fk with fk 1 and n C C reverse inequality we let f Cc (X; ) with f g h, and define j j D R fk on spt h, so the bound on the right of the previous inequality is 2 j j Ä C ! j j ( g ( h attained and we have proved g h f if g h > 0 g h f if g h > 0 f 1 C C f2 C C Z Z D 0 if g h 0; D 0 if g h 0: h d h d C D C D X D X j j 1 Then f1 f2 f , f1 g, f2 h and it is readily checked that f1; f2 and again using the density of Cc (X) in L () we have 1 -a.e. nC D j j Ä j j Ä 2 j j D Cc (X; R ). Then L(f ) L(f1) L(f2) (g) (h), and hence taking D C Ä C Math 205A Lecture Supplement 15
We conclude with an important compactness theorem for Radon Measures. Recall that Alaoglu’s theorem (see e.g. Royden, “Real Analysis” 3rd Edition, Macmillan 1988, p.237), which is a corollary of Tychonoff’s theorem, tells us that the closed unit ball in the dual space of a normed linear space must be weak* compact: that is, given any normed linear space X with dual space X (i.e. X is the normed space consisting of all the bounded linear functionals F X R), then {F X F 1} is weak* compact, meaning that for W ! 2 W k k Ä any sequence Fj X with sup Fj < there is a subsequence Fj and an 2 j k k 1 k F X with Fj (x) F (x) for each fixed x X . 2 k ! 2 In particular if X is compact and X C (X) (the continuous real-valued func- D tions on X ) X 1 is weak* compact. That is, given a se- 2 W k k Ä quence {k } of bounded linear functionals on C (X) with sup k < , k 1 k k 1 we can find a subsequence { } and bounded linear functional such that k 0 lim k (f ) (f ) for each fixed f C (X). Using the above Riesz Rep- 0 D 2 resentation 2.12, this implies the following assertion concerning sequences of Radon measures on X, assuming X is -compact.
2.15 Theorem (Compactness Theorem for Radon Measures.) Suppose {k } is a sequence of Radon measures on the locally compact, -compact Hausdorff space X with the property sup k (K) < for each compact K. Then there is a subse- k 1 { } X quence k 0 which converges to a Radon measure on in the sense that Z Z lim f d f d; f C (X): k 0 for each c X D X 2
Proof: Let K1; K2;::: be an increasing sequence of compact sets with X R D j Kj and let Fj;k C (Kj ) R be defined by Fj;k (f ) f dk, k [ W ! D Kj D 1; 2; : : :. By the Alaoglu theorem there is a subsequence F and a non- j;k 0 negative bounded functional Fj C (Kj ) R with Fj;k (f ) Fj (f ) for W ! 0 ! each f C (Kj ). By choosing the subsequences successively and taking a 2 diagonal sequence we then get a subsequence k and a non-negative linear R 0 F Cc (X) R with f dk F (f ) for each f Cc (X), where W ! X 0 ! 2 F (f ) Fj (f Kj ) whenever spt f Kj . (Notice that this is unambiguous D j because if spt f Kj and ` > j then F`(f K`) Fj (f Kj ) by construction.) j D j Then by applying Theorem 2.12 we have a Radon measure on X such that R R R F (f ) f d for each f Cc (X), and so f dk f d for each D X 2 X 0 ! X f Cc (X). 2